The Natural Numbers are written successively as shown below:
12345678910111213141516..........., such that the 4th digit is '4' the 9th digit is '9' but the 11th digit is '0', the 15th digit '2', the 17th '3', and so on.
What is the 40,000th digit that appears in this list ?

(In reply to Solution by dennis wenne)

3 mistakes you made:

1. There are 9 1-digit-numbers which are relevant for this problem cause 0 is left out. Which means that we have 38889 digits in front of the first 5-digit-number.

2. 40,000-38889 = 1111

3. Thus we need the first digit after the 222nd 5-digit-number like the other solutions tell you. Your last mistake was that 10,022 is not the 22nd 5-digit-number. It's the 23rd.
But we need the 223rd which is 10,222 and the first digit is 1.
Edited on September 15, 2003, 7:11 pm

Posted by abc on 2003-09-15 18:55:27